Astronomy Principles and Practice Fourth Edition.pdf

220 The radiation laws
The brightness of a first magnitude star is, therefore, 100 times greater than that of a sixth
magnitude star. This is a very useful figure to remember.
Example 15.2. When a telescope is pointed to two stars in turn, the received power is 5·3 × 10 −14 W
and 3·9 × 10 −14 W. What is the difference in apparent magnitude of these stars
Now the energy received is proportional to brightness. Therefore, by equation (15.11),
( ) 5·3
m 1 − m 2 = −2·5log 10
3·9
= −2·5 × 0·13
= −0·325.
Note that, in this case, the magnitude difference (m 1 − m 2 ) is a negative quantity, indicating that
m 2 > m 1 , this being so since B 1 > B 2 .
Example 15.3. At an observing site the brightness of the night sky background per ¾ ′′ is equivalent
to a 21st magnitude star. In the search for a faint object a telescope scans the sky with a field of view
limited to 200 ¾ ′′ . Calculate the equivalent magnitude of the star matching the total effective brightness
of the sky background.
The ratio of recorded energy in the experiment to that from 1 ¾ ′′ is 200:1. Using equation (15.11)
m 200 − 21 =−2·5log 10 ( 200
1 )
giving m 200 = 21 − 5·75 = 15·m25.
15.7 Spectral lines
15.7.1 Introduction
Soon after the turn of the 20th century, experiments were performed which revealed that atoms
had component particles. As a result of the investigations of electrical discharges through gases,
a ‘radiation’ was discovered which caused a fluorescence on the walls of the glass discharge tube
opposite the end at which the negative voltage was applied. The radiation was at first given the name
cathode rays, as it appeared to emanate from the negative terminal or cathode. Experiments with
electric and magnetic fields demonstrated that the rays consisted of negatively charged particles and
the name electron was given to them. Determination of the ratio of their charge to their mass (e/m e )
showed that it was about 1840 times the same ratio (e/M) obtained by Faraday for the hydrogen ion.
As the charges on the two types of particle were found to be of the same value (but different signs),
it follows that the mass of the electron is only 1/1840 times the mass of the hydrogen ion. It was
immediately obvious that with such a low mass, the electron could not take its place in the periodic
table of chemical elements and it was suggested that it constituted one of the fundamental parts of an
atom. An experiment by Millikan in 1905 gave a measure of electronic charge (1·6 × 10 −19 C) and
this allowed determination of the mass m e of the electron (9·1 × 10 −31 kg).
In a further experiment with the discharge tube, Goldstein punctured the cathode with small holes
and discovered what he called canal rays which appeared to flow from the anode of the tube. Again
these rays were found to have a particle nature but the e/M ratio for the particles depended on the
gas contained in the tube. The highest value for e/M is obtained when the canal rays are produced
in a hydrogen discharge tube. This was indicative of the hydrogen ion being the fundamental unit of
positive charge and it was, therefore, called a proton.
With the discovery of two of the fundamental particles—the electron and the proton—the problem
of understanding the nature of atoms was tackled. A further experiment by Lord Rutherford in 1911,